A 20 volt battery has a 4 ohm internal resistance. There is a 5 amp fuse in the main line of the battery. How many 6 ohm resistors can be connected in parallel across the battery before thefuse blows?

Answer 1

This is a trick question. If the external load is x ohms, then the fraction of the voltage drop across the load will be x/(x+4), or 20x/(x+4) volts.

You can see that if the external load is 0, then the current is 5 amps. But, that would take an infinite number of resistors in parallel.

Answer 2

Summing up:
I= 5 A
V=20 V

From V=IR, we have
R=V/I=20/5=4 Ω
So without anything in parallel with the battery, it is at the limit of tripping the fuse.
Any number of resistors will reduce the main current below 5A.

Answer 3

I bet it was 0.4 ohm internal resistance.

Answer 4

Can someone break this down a little better? Im so confused